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I am trying to design the network graph for an electrical circuit. I have done this by hand so far. Here is the code for the voltages graph

g = Graph[{0, 1, 2, 3, 4},
   {0 -> 1, 1 -> 2, 2 -> 1, 3 -> 2, 4 -> 3, 0 -> 4, 4 -> 0, 1 -> 3}];
PropertyValue[g, VertexLabels] = Table[i -> StringForm["(`1`)", i], {i, 0, EdgeCount[basic]}]; 

PropertyValue[{g, 0 -> 1}, EdgeLabels] = Placed[2, {1/2, {1/2, 0}}];
PropertyValue[{g, 1 -> 3}, EdgeLabels] = Placed[10, {1/2, {3/2, 1/2}}];
PropertyValue[{g, 1 -> 2}, EdgeLabels] = Placed[2, {1/2, {-1/2, -1/2}}];
PropertyValue[{g, 0 -> 4}, EdgeLabels] = Placed[3, {1/2, {-1/2, 1/2}}];

Here is the result:

enter image description here

I have a similar one for currents and now I need to apply Kirchhoff's laws and the only way out of this right now is doing it by hand. So my question is: is there any other more efficient way around what I am doing here?

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  • $\begingroup$ Please note that I have just started using Mathematica a few weeks ago! $\endgroup$ – Victor Apr 11 at 18:35
  • $\begingroup$ A circuit of resistors is not a directed graph. It is undirected. It does make sense to orient edges so that we can distinguish between currents flowing in opposite direction (there's a frame of reference), but you have edges going in both directions between the same vertices. It's not clear what you are trying to represent with that. $\endgroup$ – Szabolcs Apr 11 at 19:15
  • $\begingroup$ @Szabolcs edge 4->0 may represent a resistor (with current flowing from 4 to 0) and edge 0->4 may represent a voltage source. $\endgroup$ – Victor Apr 11 at 19:27
  • $\begingroup$ @Szabolcs although not in English, please take a look at the two diagrams in the middle of page 8 of this document. That's an example of what I have to do here, or the figures at the top of page 23 in this document $\endgroup$ – Victor Apr 11 at 19:29
  • 2
    $\begingroup$ "Sagetile de pe laturi indica sensurile referinta ale curentilor si tensiunilor", thus the edge directions do not have physical meaning, they only serve as a reference for the current values and voltage difference values. (I can read Romanian.) $\endgroup$ – Szabolcs Apr 11 at 20:02
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A resistor network can be represented with an undirected (multi-)graph. We are going to orient each edge to obtain a directed graph, so we have a reference for which direction the current is flowing.

Let $B$ be the vertex-edge incidence matrix of the oriented graph. This can be obtained with IncidenceMatrix.

Let $v=(v_1, \dots, v_n)$ be the vector of voltages at each node, $j=(j_1, \dots, j_m)$ the vector of currents through each edge, and $c = (c_1, \dots, c_m)$ the conductance of each resistor.

Let us put a voltage between nodes $s$ and $t$.

Kirchoff's current law tells us that the sum of currents is zero at each node except $s$ and $t$ where it is some $i$ and $-i$ respectively. In matrix notation, the sum of currents at each node is $Bj$.

Ohm's law tells us that $j = c (B^T v)$.

Putting the two together we get the sum of currents at each node as $B C B^T v$ where $C$ is a diagonal matrix obtained from $c$.

Now in Mathematica,

edges = {
   1 -> 2,
   1 -> 2,
   1 -> 3,
   2 -> 4,
   4 -> 3,
   5 -> 6,
   6 -> 4,
   5 -> 1
   };

SeedRandom[42];
conductances = RandomReal[{0.1, 1}, Length[edges]]

g = Graph[edges]

enter image description here

b = IncidenceMatrix[g]

c = DiagonalMatrix@SparseArray[conductances]

s = 1; t = 6; (* index of sink and source node *)
totalCurrent = 1 (* total current from s to t *)

Now we can get the voltages at each node.

voltages = 
 LinearSolve[
   b.c.Transpose[b],
   ReplacePart[
     ConstantArray[0, VertexCount[g]],
     {s -> -totalCurrent, t -> totalCurrent}
   ]
 ]

This system is underdetermined (corresponding to the fact that there's no reference for the voltages and only voltage differences make sense), but luckily Mathematica is smart enough to deal with that.

Get the current through each edge:

currents = conductances (voltages.b)

Get the effective resistance between s and t:

effectiveResistance = (voltages[[t]] - voltages[[s]])/current

Unfortunately, Mathematica is not capable of styling parallel edges differently. Below I'll use a simple graph (no multi-edges) to illustrate how to visualize the result.

Let this be our graph:

ug = Graph[GraphData["GreatRhombicuboctahedralGraph"], 
   GraphStyle -> "BasicBlack", VertexSize -> 0.5];

We orient edges in an arbitrary way:

g = DirectedGraph[ug, "Acyclic"]

Then use the above code, but set the same conductance for all edges,

conductances = N@ConstantArray[1, EdgeCount[g]];

and choose

s = 1; t = 12;

Visualize voltages:

Graph[
 ug,
 VertexStyle -> 
    Thread[VertexList[ug] -> ColorData["Rainbow"] /@ Rescale[voltages]],
 VertexSize -> {s -> 1, t -> 1}
]

enter image description here

Visualize current magnitudes though each edge:

Graph[ug,
 EdgeStyle -> Prepend[
   Thread[EdgeList[ug] -> (ColorData["Rainbow"] /@ Rescale@Abs[currents])],
   Thickness[0.02]],
 VertexSize -> {s -> 1, t -> 1}
]

enter image description here

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  • $\begingroup$ This is really clever! Thank you so much! One question: you added two edges 1 -> 2 in the edges list. Is that correct? $\endgroup$ – Victor Apr 11 at 20:07
  • $\begingroup$ @Victor Yes, I did that on purpose to illustrate that the method works even with multigraphs. These two edges represent two resistors in parallel. In general, Mathematica does not deal well with multigraphs, so it's good to know that there's no problem here (IncidenceMatrix can handle them). Visualization will be troublesome though, so I added a non-multigraph example for that. $\endgroup$ – Szabolcs Apr 11 at 20:15
  • $\begingroup$ Thank you so much! It became clearer! $\endgroup$ – Victor Apr 11 at 20:16

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